# Logic equivalence question

I have a question about discrete mathematics question that I have been struggling to solve. Here is the question:

Each of the two rooms (room I and room II) contains either a lady or a tiger. If a room contains a lady, the sign on its door is true. If it contains a tiger, the sign is false. The signs are Room 1 - It makes no difference which room to pick. Room 2 - There is a lady in the other room. which room contains ladies? Use logic equivalences or rules of inference.

The only steps I could achieve were: I let p - there is a lady in room 1 q - there is a lady in room 2 and I said that sign 1 represents: (p ^ q) ∨ (~p ^ ~q) and I also said sign 2 represents: q ↔ p I don't know if it's correct but please provide your guidance. Thank you🙏

• The stakes certainly are high, aren't they? What are your thoughts/attempts? Sep 29 at 4:01
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Sep 29 at 4:01
• Just so you know: people on this site aren't interested in giving answers you can copy and paste to get marks from your teacher. However they are interested in working with you to help you find an answer and learn about mathematics. To start this process, please include in your post any ideas or thoughts you have. Sep 29 at 4:11
• Wouldn't the sign on room 2 simply be $p$? Sep 29 at 4:38
• Now translate the conditions on whether the signs are true: "If room 1 contains a lady, then sign 1 is true" becomes "$p \to ((p \wedge q) \vee (\sim p \wedge \sim q))$," etc. Sep 29 at 4:40

You got a good start.

Yes, sign 2 is saying $$p$$, and since the sign is true if and only if there is a lady in it, the information that you have with regard to room 2 is indeed: $$q \leftrightarrow p$$

For room 1 you made a small mistake. Yes, you got the right idea for symbolizing what the sign for room 1 is saying, which is that either both rooms contain ladies or both rooms contain tigers, which you symbolized correctly as $$(p \land q) \lor (\neg p \land \neg q)$$. However, again you need to use the information that what the sign is saying is true if and only if there is a lady in the room, meaning that the information you have with regard to room 1 is: $$p \leftrightarrow ((p \land q) \lor (\neg p \land \neg q))$$

OK, so now you need to combine these two pieces of information to get the result. (Hint: think of what makes $$q \leftrightarrow p$$ true). Good luck!

• Thank you! You explained it very well. Oct 4 at 2:34
• @RosePink You're welcome! I assumed you figured it out? Oct 4 at 13:01
• yes! I figured out. Oct 5 at 23:38

the trick is to take 2 cases that lady is in room $$1$$ or $$2$$, and then follow the arguments given in the question for both cases, and arrive at a contradiction in any one of the cases...

using propositional logic ( its enough )

let $$r_1$$ be room $$1$$ and $$r_2$$ be room $$2$$.

Let $$\mathbf L \$$ be lady $$\ \$$ and $$\ \ \ S_1 , S_2$$ be signs on room $$1$$ and room $$2$$ respectively

let $$\mathbf P =$$ L is in $$r_1$$ ,$$\ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \$$ let $$\mathbf Q =$$ L is in $$r_2$$

let $$\mathbf A =$$ $$S_1$$ is true,$$\ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \$$ let $$\mathbf B =$$ $$S_2$$ is true

its given that $$( P \rightarrow \lnot Q ) \ \land \ ( Q \rightarrow \lnot P )$$ .... $$(a)$$

now, $$\ \ \ \bf CASE \ \ 1$$

$$(1)$$ $$\quad$$ if $$\quad$$ Q is true ,

$$(2)$$ then $$\ \ \ \$$ Q $$\rightarrow$$ B, $$\ \ \ \therefore S_2$$ is true

$$(3)$$ but its given that $$B \rightarrow P$$, $$\ \ \ \therefore P$$ is true

$$\therefore \ \text{from} \ \ (a), (1) \ \& \ (3)$$ .. $$\bf Q$$ leads to a contradiction... $$AND \ \ \ \bf P$$ must be $$\bf TRUE$$.

( btw, you already knew lady was in room$$1$$, I believe )